'Weak Dependency Graph [60.0]'
------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    innermost runtime-complexity with respect to
  Rules:
    {  a(c(x1)) -> c(b(x1))
     , a(x1) -> b(b(b(x1)))
     , b(c(b(x1))) -> a(c(x1))}

Details:         
  We have computed the following set of weak (innermost) dependency pairs:
   {  a^#(c(x1)) -> c_0(b^#(x1))
    , a^#(x1) -> c_1(b^#(b(b(x1))))
    , b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
  
  The usable rules are:
   {  b(c(b(x1))) -> a(c(x1))
    , a(c(x1)) -> c(b(x1))
    , a(x1) -> b(b(b(x1)))}
  
  The estimated dependency graph contains the following edges:
   {a^#(c(x1)) -> c_0(b^#(x1))}
     ==> {b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
   {a^#(x1) -> c_1(b^#(b(b(x1))))}
     ==> {b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
   {b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
     ==> {a^#(x1) -> c_1(b^#(b(b(x1))))}
   {b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
     ==> {a^#(c(x1)) -> c_0(b^#(x1))}
  
  We consider the following path(s):
   1) {  a^#(c(x1)) -> c_0(b^#(x1))
       , b^#(c(b(x1))) -> c_2(a^#(c(x1)))
       , a^#(x1) -> c_1(b^#(b(b(x1))))}
      
      The usable rules for this path are the following:
      {  b(c(b(x1))) -> a(c(x1))
       , a(c(x1)) -> c(b(x1))
       , a(x1) -> b(b(b(x1)))}
      
        We have applied the subprocessor on the union of usable rules and weak (innermost) dependency pairs.
        
          'Weight Gap Principle'
          ----------------------
          Answer:           YES(?,O(n^1))
          Input Problem:    innermost runtime-complexity with respect to
            Rules:
              {  b(c(b(x1))) -> a(c(x1))
               , a(c(x1)) -> c(b(x1))
               , a(x1) -> b(b(b(x1)))
               , a^#(c(x1)) -> c_0(b^#(x1))
               , b^#(c(b(x1))) -> c_2(a^#(c(x1)))
               , a^#(x1) -> c_1(b^#(b(b(x1))))}
          
          Details:         
            We apply the weight gap principle, strictly orienting the rules
            {  a(c(x1)) -> c(b(x1))
             , a(x1) -> b(b(b(x1)))}
            and weakly orienting the rules
            {}
            using the following strongly linear interpretation:
              Processor 'Matrix Interpretation' oriented the following rules strictly:
              
              {  a(c(x1)) -> c(b(x1))
               , a(x1) -> b(b(b(x1)))}
              
              Details:
                 Interpretation Functions:
                  a(x1) = [1] x1 + [1]
                  c(x1) = [1] x1 + [0]
                  b(x1) = [1] x1 + [0]
                  a^#(x1) = [1] x1 + [1]
                  c_0(x1) = [1] x1 + [1]
                  b^#(x1) = [1] x1 + [0]
                  c_1(x1) = [1] x1 + [1]
                  c_2(x1) = [1] x1 + [0]
              
            Finally we apply the subprocessor
            We apply the weight gap principle, strictly orienting the rules
            {a^#(x1) -> c_1(b^#(b(b(x1))))}
            and weakly orienting the rules
            {  a(c(x1)) -> c(b(x1))
             , a(x1) -> b(b(b(x1)))}
            using the following strongly linear interpretation:
              Processor 'Matrix Interpretation' oriented the following rules strictly:
              
              {a^#(x1) -> c_1(b^#(b(b(x1))))}
              
              Details:
                 Interpretation Functions:
                  a(x1) = [1] x1 + [1]
                  c(x1) = [1] x1 + [0]
                  b(x1) = [1] x1 + [0]
                  a^#(x1) = [1] x1 + [1]
                  c_0(x1) = [1] x1 + [1]
                  b^#(x1) = [1] x1 + [0]
                  c_1(x1) = [1] x1 + [0]
                  c_2(x1) = [1] x1 + [0]
              
            Finally we apply the subprocessor
            We apply the weight gap principle, strictly orienting the rules
            {a^#(c(x1)) -> c_0(b^#(x1))}
            and weakly orienting the rules
            {  a^#(x1) -> c_1(b^#(b(b(x1))))
             , a(c(x1)) -> c(b(x1))
             , a(x1) -> b(b(b(x1)))}
            using the following strongly linear interpretation:
              Processor 'Matrix Interpretation' oriented the following rules strictly:
              
              {a^#(c(x1)) -> c_0(b^#(x1))}
              
              Details:
                 Interpretation Functions:
                  a(x1) = [1] x1 + [1]
                  c(x1) = [1] x1 + [0]
                  b(x1) = [1] x1 + [0]
                  a^#(x1) = [1] x1 + [1]
                  c_0(x1) = [1] x1 + [0]
                  b^#(x1) = [1] x1 + [0]
                  c_1(x1) = [1] x1 + [1]
                  c_2(x1) = [1] x1 + [0]
              
            Finally we apply the subprocessor
            'fastest of 'combine', 'Bounds with default enrichment', 'Bounds with default enrichment''
            ------------------------------------------------------------------------------------------
            Answer:           YES(?,O(n^1))
            Input Problem:    innermost relative runtime-complexity with respect to
              Strict Rules:
                {  b(c(b(x1))) -> a(c(x1))
                 , b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
              Weak Rules:
                {  a^#(c(x1)) -> c_0(b^#(x1))
                 , a^#(x1) -> c_1(b^#(b(b(x1))))
                 , a(c(x1)) -> c(b(x1))
                 , a(x1) -> b(b(b(x1)))}
            
            Details:         
              The problem was solved by processor 'Bounds with default enrichment':
              'Bounds with default enrichment'
              --------------------------------
              Answer:           YES(?,O(n^1))
              Input Problem:    innermost relative runtime-complexity with respect to
                Strict Rules:
                  {  b(c(b(x1))) -> a(c(x1))
                   , b^#(c(b(x1))) -> c_2(a^#(c(x1)))}
                Weak Rules:
                  {  a^#(c(x1)) -> c_0(b^#(x1))
                   , a^#(x1) -> c_1(b^#(b(b(x1))))
                   , a(c(x1)) -> c(b(x1))
                   , a(x1) -> b(b(b(x1)))}
              
              Details:         
                The problem is Match-bounded by 0.
                The enriched problem is compatible with the following automaton:
                {  c_0(2) -> 2
                 , b_0(2) -> 9
                 , b_0(9) -> 8
                 , a^#_0(2) -> 4
                 , c_0_0(6) -> 4
                 , b^#_0(2) -> 6
                 , b^#_0(8) -> 7
                 , c_1_0(7) -> 4}